Solution Day 14, part two

The easter egg configuration is the first with only unique robot positions
(a. e. every position in the grid is occupied by at most one robot).

Author: rhaendel

src/main/kotlin/Day14.kt

@@ -1,5 +1,3 @@
-import de.ronny_h.extensions.isSymmetrical
-
 fun main() {
     val day = "Day14"
 
@@ -103,50 +101,28 @@ fun main() {
 
     println("$day part 2")
 
-    fun List<Robot>.isNotSymmetrical(width: Int, height: Int): Boolean {
+    fun List<Robot>.isNotUnique(width: Int, height: Int): Boolean {
         val grid = toGrid(width, height)
-        val symmetrical = grid.all { row ->
-            row.isSymmetrical()
+        val allUnique = grid.all { row ->
+            row.all { it <= 1 }
         }
-        if (symmetrical) {
+        if (allUnique) {
             grid.print()
         }
-        return !symmetrical
+        return !allUnique
     }
 
     fun iterateUntilSymmetrical(robots: List<Robot>, width: Int, height: Int): Long {
         var seconds = 0L
         do {
             seconds++
-            if (seconds % 10000 == 0L) print('.')
-            if (seconds % 1000000 == 0L) println()
             val result = robots.move(width, height, seconds)
-        } while (result.isNotSymmetrical(width, height))
+        } while (result.isNotUnique(width, height))
         return seconds
     }
 
-    fun part2Small(input: List<String>) = iterateUntilSymmetrical(input.parseRobots(), 11, 7)
-
     fun part2(input: List<String>) = iterateUntilSymmetrical(input.parseRobots(), 101, 103)
 
-    printAndCheck(
-        """
-            p=0,4 v=3,-3
-            p=6,3 v=-1,-3
-            p=10,3 v=-1,2
-            p=2,0 v=2,-1
-            p=0,0 v=1,3
-            p=3,0 v=-2,-2
-            p=7,6 v=-1,-3
-            p=3,0 v=-1,-2
-            p=9,3 v=2,3
-            p=7,3 v=-1,2
-            p=2,4 v=2,-3
-            p=9,5 v=-3,-3
-        """.trimIndent().lines(),
-        ::part2Small, 2
-    )
-
     printAndCheck(input, ::part2, 91649162)
 }